$\overline{AC}$ is $10$ units long $\overline{BC}$ is $8$ units long $\overline{AB}$ is $2\sqrt{41}$ units long What is $\sin(\angle BAC)$ ? $A$ $C$ $B$ $10$ $8$ $2\sqrt{41}$
Answer: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{BC} = 8$ hypotenuse $= \overline{AB} = 2\sqrt{41}$ $\sin(\angle BAC)=\frac{8}{2\sqrt{41}}$ $=\dfrac{4\sqrt{41} }{41}$